Harmonic Mitigation Selection

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1

Harmonic Mitigation Selection

This paper is a review of the methods available to reduce line current harmonics seen with AC motordrives. There are several methods that can reduce the Ithd, and by doing so, also reduce the Vthd on

the secondary terminals of the transformer feeding the drive. However, each one impacts the power

distribution system in a different way, and each one impacts the drive operation in a different way.This paper will review the effects along with the advantages and the disadvantages of each method.Included in this paper are test results of several of the harmonic mitigation methods using a 100hp

1336F drive.

1. Introduction to Harmonics

2. The Basic Drive System

3. AC Line Reactor4. DC Link Choke

5. Pseudo 12-Pulse

6. Multi-Pulse Converter

7. Passive Filter8. Active Filter

9. Active Front-End

10. Other Comparisons11. Comparison Checklist

12. Summary Rules-of-Thumb

Rick Hoadley

April 2005

Acknowledgments: Special thanks to Nick Guskov, Gary Woltersdorf and John Streicher for

engineering, technical and marketing assistance.



2

Harmonic Mitigation Selection

When faced with the task of reducing line current harmonics caused by AC drives, there are severalalternatives available. How does one pick the best solution? What criteria should be used to weigh

the alternatives? This paper will deal with these topics.

1. Introduction to Harmonics

ConverterAC to DC

DC BusFilter

InverterDC to AC

AC Drive

A C

L i n e

I n p u t

A C

M o t or O u t p u t

Why do drives create current harmonics?

We know that an AC drive creates line currentharmonics because the line current does not look

like a sine wave. Why? Because the converter

within the drive takes the 3 phase AC line voltagesand converts them into DC that feeds a filter and the

inverter section. The inverter section is what then

takes the DC and creates the variable voltage,

variable frequency output for the motor. See Fig 1. Fig 1. Sections of a 6-Pulse AC drive

During the process of 3-phase AC to DC conversion,

the line current can resemble a pair of distinctpulses. VMab.V = f...

La.I = f(t, b...

Cbus.V = f(...

VMac.V = f...

VMan.V [V]...

81.50m

81.50m

93.50m

93.50m

82.50m

82.50m

85.00m

85.00m

87.50m

87.50m

90.00m

90.00m

92.50m

92.50m

100.0 -100.0

00.0

0 0

00.0 200.0

00.0 400.0

00.0 600.0

Vab Vac

Van

Vbus

Ia

VMab.V = f...

Cbus.V = f(...

VMac.V = f...

VMan.V [V]...

81.50

La.I = f(t, b...

m

81.50m

93.50m

93.50m

82.50m

82.50m

85.00m

85.00m

87.50m

87.50m

90.00m

90.00m

92.50m

92.50m

100.0 -100.0

00.0

0 0

00.0 200.0

00.0 400.0

00.0 600.0

Vab Vac

Van

Vbus

Ia

Vab Vac

Van

Vbus

Ia

Why does the current look like that? Remember,

current can only flow whenever the instantaneousline-to-line voltage exceeds the DC bus capacitor

voltage. So for phase A, when can current flow?

Only when the peak voltage between phases A and

B exceed the cap voltage, or only when the peak voltage between phases A and C exceed the cap

voltage. No current flows in phase A due to thepeak voltage between phases B and C. This is

easily seen in Fig 2.

VMab.V = f...The currents in phases B and C are the same, simply

phase shifted with respect to each other.La.I = f(t, b...

Cbus.V = f(...

VMac.V = f...

VMan.V [V]...

181.5m

181.5m

193.5m

193.5m

182.5m

182.5m

185.0m

185.0m

187.5m

187.5m

190.0m

190.0m

192.5m

192.5m

100.0 -100.0

00.0

0 0

00.0 200.0

00.0 400.0

00.0 600.0

Vab Vac

Van

Vbus

Ia

VMab.V = f...

Cbus.V = f(...

VMac.V = f...

VMan.V [V]...

181.5

La.I = f(t, b...

m

181.5m

193.5m

193.5m

182.5m

182.5m

185.0m

185.0m

187.5m

187.5m

190.0m

190.0m

192.5m

192.5m

100.0 -100.0

00.0

0 0

00.0 200.0

00.0 400.0

00.0 600.0

Vab Vac

Van

Vbus

Ia

Fig 3. Current in phase A with an increase inline impedance

Fig 2. Current in phase A with very little lineimpedance

If impedance is added to the circuit, either byadding a DC link choke, or an AC line reactor, the

current does not change as rapidly, causing thewaveform to change shape as seen in Fig 3.

What causes voltage distortion?

Since the current looks like these pulses and do notlook like a sine wave, when we analyze the current

waveform we will see that it is composed of afundamental sine wave along with a combination of

several harmonics. These are the harmonic currents



3

that cause voltage distortions on the powerdistribution system.

Rfund.V =...

Rharm.V ...

Rtotal.V =...

0

0

40.00

40.00

10.00

10.00

20.00

20.00

30.00m

30.00m

150.0 -150.

50.0

0

100.0 -100.

50.0 -50.0

0.0 50.0

00.0 100.0

The voltage distortions appear whenever currentharmonics flow through the impedance within the

transformer feeding the drive, and through anyimpedance in the wiring between the transformerand the drive. As you may realize, the voltage

distortion is least at the transformer terminals, and

greatest at the drive terminals. It is this voltage

distortion that causes problems for other pieces of equipment fed by that transformer.

Fig 4. The red line current is made up of thegreen fundamental current plus the blueharmonic currents

The next two figures show how the voltagedistortion will change depending on the size of the

transformer feeding a drive. Shown in Fig 5 is the

voltage waveform at the terminals of a 1500kVAtransformer, and the current waveform of a 75hp

AC drive. There is very little distortion in the line-

to-line voltage. However, Fig 6 shows the

waveforms at the terminals of a 75kVA transformer.There is a significant amount of flat-topping in the

distorted voltage. This amounts to about 10% Vthd.Fig 5. The red line current of a 75hp drivecauses little distortion seen in the blue line-to-line voltage at the 1500kVA xfmr sec terminals

How much is too much?

The question we now ask is, “How much is toomuch?” This is where the recommendations found

in IEEE std 519-1992 come into play. Table 10.2

recommends that the voltage distortion be less than3% for hospitals and airports, less than 5% for all

other facilities, but may be as high as 10% if

converter loads are the only loads on a particulartransformer. This is shown in Fig 7. The location

for measuring the voltage distortion can be

anywhere in the facility where non-linear and linear

loads are connected together, typically at thesecondary terminals of distribution transformers. Fig 6. The red line current of a 75hp drive

causes significant distortion seen in the blueline-to-line voltage at the 75kVA xfmr sec

terminals

Why is there a concern for limiting the voltagedistortion? When motors are fed line voltages that

contain voltage distortion, there will be additional

heating in the motors and a loss of torque. It isdifficult to predict how much the motor needs to be

derated due to the level and content of the voltage

harmonics present. However, in NEMA MG1-1998,Revision 1, Part 30, Section 30.1.2.1, they define an

Harmonic Voltage Factor and a give a derating

chart. This was part of the consideration that IEEE

Application Maximum THD (%)

Special Applications - hospitals and airports 3.0%

General System 5.0%

Dedicated System - exclus ively converte r load 10.0%

Harmonic Voltage LimitsLow-Voltage Systems

Table 10.2

Application Maximum THD (%)

Special Applications - hospitals and airports 3.0%

General System 5.0%

Dedicated System - exclus ively converte r load 10.0%

Harmonic Voltage LimitsLow-Voltage Systems

Table 10.2

Fig 7. Table 10.2 from IEEE 519 showingrecommended Vthd limits for variousapplications



4

used when determining the values in Table 10.2. Inaddition to overheating motors, other equipment can

also malfunction due to voltage distortions on the

power lines.

Table 10.3

Current distortion Limits for General Distribution Systems (120V through 69,000V)

Isc/ Iload <11 11<=h<17 17<=h<23 23<=h<35 35<=h TDD (%)

<20 4.0 2.0 1.5 0.6 0.3 5.0

20<50 7.0 3.5 2.5 1.0 0.5 8.0

50<100 10.0 4.5 4.0 1.5 0.7 12.0

100<1000 12.0 5.5 5.0 2.0 1.0 15.0

>1000 15.0 7.0 6.0 2.5 1.4 20.0

Even harmonics are limited to 25% of the odd harmonic limits above

Isc=maximum short circuit current at PCCIload=maximum demand load current (fundamental frequency component) at PCC

Maximum Harmonic Current Distortion in Percent of Iload

Table 10.3

Current distortion Limits for General Distribution Systems (120V through 69,000V)

Isc/ Iload <11 11<=h<17 17<=h<23 23<=h<35 35<=h TDD (%)

<20 4.0 2.0 1.5 0.6 0.3 5.0

20<50 7.0 3.5 2.5 1.0 0.5 8.0

50<100 10.0 4.5 4.0 1.5 0.7 12.0

100<1000 12.0 5.5 5.0 2.0 1.0 15.0

>1000 15.0 7.0 6.0 2.5 1.4 20.0

Even harmonics are limited to 25% of the odd harmonic limits above

Isc=maximum short circuit current at PCCIload=maximum demand load current (fundamental frequency component) at PCC

Maximum Harmonic Current Distortion in Percent of Iload

Table 10.3 puts limits on the current harmonics atthe place where multiple customers are connected tothe utility, usually at the metering point. See Fig 8.

This is the primary PCC (Point of Common

Coupling). Table 10.3 allows each customer

connected to the utility to draw some harmoniccurrent and thus create some voltage distortion

which everyone else needs to live with. However,

no one customer should draw so much harmoniccurrent so as to cause the voltage distortion at this

common connection point to exceed the

recommended limits at this point and causeproblems for the other customers.

Fig 8. Table 10.3 from IEEE 519 showingrecommended Ithd limits for various Isc/Iloadcategories

Within a customer’s plant, though, Table 10.3 does

not apply. Instead, only Table 10.2 applies –voltage distortion limits. The user should plan his

equipment type and placement and wiring to make

sure that the voltage distortion at any transformersecondary and at any point where linear and non-

linear loads are connected together are within thelimits in Table 10.2.

Fig 9. Diagram showing overview of whatvoltage distortion limits should be followed atvarious locations in multiple systems

If drives or other converters are the only load on atransformer, this table allows the voltage distortion

to be as high as 10%! For most users, the voltage

distortion should be less than 5%. For hospitals andairports (life-critical applications), the voltage

distortion should not exceed 3%. This is

diagrammed in Fig 9.

If, after a survey and harmonic estimation is made

of the plant, it is determined that there is a need to

reduce the harmonic current drawn by a drive or agroup of drives, several methods are available. We

will review these methods, compare their ability to

reduce the line current harmonics, list the pros andcons of each one, and discuss the effect each

method has on the operation of the drive and on the

power distribution system.

We will start with the characteristics of a basic,

bare-bones drive and draw comparisons from that.



5

2. The Basic Drive System Transformer

xfmr% Z

Drive

DC

AC

AC

DCM

The basic drive system, in Fig 10, consists of a

transformer as the power source, a 6-pulse diodebridge converter, a DC bus filter capacitor, an IGBT

inverter bridge, and a motor connected to amechanical load. Fig 1 shows the details of thedrive. The user needs to realize that the transformer

is as much a part of the system, and has an impact

on the harmonics produced by the drive.

Fig 10. The basic drive system, with atransformer feeding an unbuffered drive (a drive

without a DC link choke)

La.I = f( ...

150.0

150.0

200.0

200.0

162.5

162.5

175.0

175.0

187.5

187.5

00.0 -400.

00.0

0 0

00.0 -200.

00.0 200.0

The line current will typically look like what is

shown in Fig 11.

Analyzing this current waveform, we see that it

contains the harmonics shown in Fig 12. The %

Ithd is typically 80 to 120%, meaning that for a100hp drive, with a fundamental current of about

100Arms, the harmonic current content could be

between 80 and 120Arms! Please note that the onlyimpedance limiting the rate of rise of the current

pulses is the transformer impedance and any other

impedance that exists in the cabling between the

transformer and the drive.

Fig 11. Typical line current for this driveconfiguration

0

10

20

30

40

50

60

70

80

90

100

0 6 12 1 8 24 30 36 42 4 8 54 60 66 72 78

Harmonic Number

% H a r m o n i c C u r r e n

Most drives that are rated 5hp or less are like thisbasic drive. One or two drives in a plant usually

don’t cause any problems, but if you have 200*5hp

drives, this is the equivalent to a 1000hp drive. Thepeak currents can be around 3000Apk – causing

heating problems in the transformer and wiring,

causing circuit breakers to open or fuses to blow,and greatly distorting the voltage on the secondary

of the distribution transformer that could cause

other equipment connected to the transformer tomalfunction.

Fig 12. Harmonic spectrum showing themagnitudes of the individual harmonicscontained in the line current

30.00

40.00

50.00

60.00

70.00

80.00

90.00

100.00

0 10 20 30 40 50 60 70 80 90 100

PF disp

PF dist

PF total

Simply adding an AC line reactor, as discussed in

the next section, will eliminate a lot of the problemsassociated with this drive configuration.

Fig 13 shows how the power factor is affected bythe harmonics as the load is varied from 0 to 100%.

The green trace is the displacement power factor,

the blue trace is the distortion power factor, and thepink trace is the total power factor (the product of

the other two).

Fig 13. Power factor vs load - green isdisplacement PF, blue is distortion PF, pink istotal PF



6

3. AC Line Reactor Transformer

xfmr% Z

Line Reactor

Drive

DC

AC

AC

DCM

Adding an AC line reactor to the drive, as shown in

Fig 14 and 15, has a dramatic effect in reducing theline current harmonics. The waveform in Fig 16

shows peak currents that are about half of what wasseen in the previous configuration. The % Ithd isnow typically 30 to 60%. Fig 17 shows the

harmonic spectrum. Notice how the 5th and 7th

harmonic currents have dropped.

Fig 14. Drive configuration with the addition ofAC line reactors

ConverterAC to DC

InverterDC to AC

DCBusFilter

AC Drive

A C


A C

L i n e

I n p u t

Not only does the AC line reactor reduce the line

current harmonics, but it also helps to prevent the

drive from tripping as a result of power line voltagedips and the voltage transients that occur when

power factor correction caps are connected to the

power system. Fig 15. Details of drive configuration with ACline reactors

Fig 18 shows what typically happens during a

power dip. When the line voltage drops to half of

nominal for only a portion of a cycle, the DC busvoltage starts to drop quickly because the drive is

providing power to the motor during this time and

the energy from the capacitors is not beingreplenished from the power line. When the line

voltage returns to normal, there is a large surge of current “recharging” the DC bus caps. The DC bus

then quickly shoots up. This recharging current

surge can damage the diode bridge. Simplyinserting a 3% line reactor in the circuit will greatly

reduce the peak recharging current as seen in Fig 19.

This can prevent nuisance fuse blowing.

La.I = f(t...

150.0m

150.0m

200.0m

200.0m

162.5m

162.5m

175.0

175.0

187.5m

187.5m

00.0 -400.0

00.0

0 0

200.0 -200.0

00.0 200.0

Fig 16. Typical line current for this driveconfiguration

0

10

20

30

40

50

60

70

80

90

100

0 6 12 18 24 30 36 42 48 54 60 66 72 78

Harmonic Number


An AC line reactor also helps prevent OverVoltage

trips whenever power factor correction caps areswitched onto the power lines. The voltage

waveform that normally results during this time is

seen in Fig 20, showing how the DC bus voltage in

the drive quickly tries to charge up to the peak of the ringing waveform. Again, a large surge of

current can also result. In addition, the DC bus

voltage can be taken to a level that will cause anOverVoltage trip. Inserting a 3% AC line reactor

limits the peak charging current and limits the effect

on the DC bus voltage as seen in Fig 21. If youhave a drive that trips on OverVoltage every

morning at 7:00am, this could be the reason.




7

The only drawback in the use of AC line reactors isthat it creates a drop in the DC bus voltage as full

load, full speed is approached. Fig 22 shows how

the DC bus voltage decreases as the load increases,comparing a basic drive to a drive with a 3% line

reactor, a drive with a 5% line reactor, and a drivewith a DC link choke. A 3% line reactor typicallycauses a 3% drop in bus voltage. A 5% line reactor

causes a 5% drop in bus voltage. The effect this has

on the drive system is that full power would not be

available from the motor because rated V/Hz couldnot be achieved from the drive. If full load, full

speed operation is not required, then this is not an

issue. However the user needs to be aware of thiseffect.

Vbus = 677Vdc

Ipk = 32A

Vbus = 587Vdc

Ipk = 120A

Vbus = 677Vdc

Ipk = 32A

Vbus = 587Vdc

Ipk = 120A

Fig 18. Operation during a power dip without anAC line reactor– green is the DC bus voltage,blue is the line-to-line voltage, red is the buscurrent

Vbus = 637Vdc

Ipk = 10A

Vbus = 545Vdc to 692Vdc

Ipk = 37A

Vbus = 637Vdc

Ipk = 10A

Vbus = 545Vdc to 692Vdc

Ipk = 37A

It is possible to place multiple drives on a singleline reactor. A rule of thumb for this is to select a

3% line reactor for 3 drives in parallel, or a 5% line

reactor for 5 drives in parallel. The hp rating of the

line reactor would be the total hp of the drivesconnected to it. Fig 23 diagrams this configuration.

Fig 19. Operation during a power dip with an ACline reactor – green is the DC bus voltage, blueis the line-to-line voltage, red is the bus current

Vbus = 677Vdc

Ipk = 32A

Vbus = 806Vdc

Ipk = 507A

Vbus = 677Vdc

Ipk = 32A

Vbus = 806Vdc

Ipk = 507A

% Vbus vs Load (Isc/Iload = 47)

94.00

95.00

96.00

97.00

98.00

99.00

100.00

101.00

102.00

0 10 20 30 40 50 60 70 80 90 100

unbuffered

buffered

3% line reactor

5% line reactor

1.35*Vac

% Vbus vs Load (Isc/Iload = 47)

94.00

95.00

96.00

97.00

98.00

99.00

100.00

101.00

102.00

0 10 20 30 40 50 60 70 80 90 100

unbuffered

buffered

3% line reactor

5% line reactor

1.35*Vac Fig 20. Operation during a ringing transientwithout an AC line reactor - green is the DC bus

Vbus = 637Vdc

Ipk = 10A

Vbus = 653Vdc

Ipk = 19A

Vbus = 637Vdc

Ipk = 10A

Vbus = 653Vdc

Ipk = 19A

Fig 22. Vbus vs Load – drop in DC bus voltageas the load is increased, showing the effect anAC line reactor has on this parameter

ge, red is thevoltage, blue is the line-to-line voltabus current

Transformer

xfmr% Z

Line Reactor

Drive

DC

AC

AC

DC

DC

AC

AC

DC

DC

AC

AC

DC

M

M

M

Fig 21. Operation during a ringing transient withan AC line reactor – green is the DC busvoltage, blue is the line-to-line voltage, red is thebus current

Fig 23. Diagram of multiple unbufferred driveson a single AC line reactor



8

4. DC Link Choke Transformer

xfmr% Z

Drive

DC

AC

AC

DC

DC LinkChoke

M

Most industrial drives larger than 5hp come

equipped with a DC link choke as seen in Fig 24and 25. It usually is made up of two windings on a

common core, with one winding in each of the DCbus legs (one in the positive leg, one in the negativeleg) between the output of the converter bridge and

the bus caps.

Fig 24. Drive configuration with the addition ofDC link chokes (a buffered drive)

ConverterAC to DC

DC BusFilter

InverterDC to AC

AC Drive

A C

L i n e

I n p u t

A C


Since this adds inductive impedance to the circuit,the rate of rise of the current into the caps is limited,

just like the effect seen by adding a line reactor to

the input to the converter bridge. There is adifference, though. During operation above about

25% load, the current in the link choke becomes

continuous (seen when the two pulses on the inputmerge into one pulse with two bumps, and the

current does not go to zero between the pairs of

pulses as compared to Fig 11). The line current will

typically look like what is shown in Fig 26. Fig 27shows the harmonic spectrum. Ithd is normally

between 30 to 40%. Since the impedance of a DC

link choke is usually selected to be about 4%, youcan see that this had the effect of further reducing

the 5th and 7th harmonics in the line current.

Fig 25. Details of drive configuration with DC

link chokes

La.I = f(t...

150.0

150.0

200.0

200.0

162.5

162.5

175.0

175.0

187.5

187.5

00.0 -400.

00.0

0 0

00.0 -200.

00.0 200.0

Fig 26. Typical line current for this drive

configurationWhen operating, the voltage drop due to the DC

link choke is due to its DC resistance and not to itsinductive impedance. The result is that the DC bus

voltage does not drop as much as it would when

compared to a line reactor. See Fig 22. This is themain advantage of using a link choke instead of line

reactors. The other advantage is that they are

usually slightly less expensive than AC line reactors

since they only have two windings. An advantageof line reactors over link chokes is that they provide

better current sharing when paralleling diode

bridges.0

10

20

30

40

50

60

70

80

90

100

0 6 12 1 8 24 30 36 42 4 8 54 60 66 72 78

Harmonic Number


What is the overall effect of a link choke on a drive

and the power grid? Fig 28 to 31 will give yousome overall operational parameters with which to

compare various harmonic mitigation solutions.


Fig 28 shows how the Arms of the total line current

varies with load. It is fairly linear. The current at

100% load is about 108Arms. This current is



9

composed of about 101Arms of fundamental currentand 40Arms of harmonic current.

0.0

20.0

40.0

60.0

80.0

100.0

120.0

0 25 50 75 100

6p

Fig 29 shows the total power factor associated withthe line current vs load. At full load, the power

factor is about 0.92 lagging, and slowly drops downto about 0.7 lagging at 10% load.

Fig 30 shows the change in the DC bus voltage as

the load is changed. 100% of nominal DC bus

voltage is taken to be 1.35*Vac line-to-line voltage.For a 480Vac drive, the nominal DC bus voltage

works out to be 648Vdc. At no load, you see that

the bus voltage will charge up to the peak of the linevoltage, or 1.41*Vac. Again, for a 480Vac drive,

the peak would be 679Vac, or 1.05*nominal.

Please notice that at 100% load, the voltage drop isvery slight, due to the voltage drop at the

transformer terminals due to the internal impedance

of the secondary windings, and due to the DC

resistance of the DC link chokes.

Fig 28. Total Line Current vs % Load

0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

1.60

1.80

2.00

0 25 50 75 100

6p

0.00

0.20

0.40

0.60

0.80

L a g

L e a d

0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

1.60

1.80

2.00

0 25 50 75 100

6p

0.00

0.20

0.40

0.60

0.80

L a g

L e a d

Fig 31 shows how Iharm (the harmonic current)

varies with load. It also shows the % Ithd as adashed line. This is the ratio of the harmonic

current to the fundamental current that is containedin the AC line current.

Fig 29. Total Power Factor vs % Load

90.0

92.0

94.0

96.0

98.0

100.0

102.0

104.0

106.0

108.0

110.0

0 25 50 75 100

6p

Fig 32 is similar to Fig 13, showing how thedisplacement power factor and distortion power

factor vary with load. The displacement power

factor is quite high throughout the load range.

30.00

40.00

50.00

60.00

70.00

80.00

90.00

100.00

0 10 20 30 40 50 60 70 80 90 100

PF disp

PF dist

PF total

Fig 30. % Nominal DC Bus Voltage vs % Load

0.0

20.0

40.0

60.0

80.0

100.0

120.0

0 20 40 60 80 100

Ithd

Iharm


Fig 31. % Ithd and Iharm vs % Load



10

5. Pseudo 12-PulseDrive

DC

AC

DC LinkChoke

AC

DCM

Drive

DC

AC

DC LinkChoke

AC

DCM

Transformer

xfmr% Z

A Pseudo 12-Pulse system is one that approaches a

true 12-pulse, but where the 5th

and 7th

harmonicsare not totally cancelled. Fig 33. This is

accomplished by placing half of the hp load on adelta-wye transformer, and the other half on a delta-delta transformer (or a simple line reactor could be

used in place of the delta-delta transformer, with the

same impedance as the delta-wye transformer).

Fig 33. Pseudo 12-Pulse configuration with halfof the hp drive load on a delta-wye xfmr, and theother half on a delta-delta xfmr or line reactor

La1.I = f...

150.0m

150.0m

200.0m

200.0m

162.5m

162.5m

175.0m

175.0m

187.5m

187.5m

-1000.0 -1000.0

1000.0

0 0

-500.0 -500.0

500.0 500.0

The cancellation of the lower order harmonics will

be affected by the amount of load on each set of

drives. Any pre-existing harmonics present on thevoltage feeding these transformers will also affect

the harmonic cancellation.

Fig 34 shows the secondary current from the main

transformer when the drive on the delta-delta

transformer is at 100% load, the drive on the delta-

wye transformer is at 100% load.

Fig 34. Line current with 100% load on delta-delta and 100% load on delta-wye

La1.I = f...

150.0m

150.0m

200.0m

200.0m

162.5m

162.5m

175.0m

175.0m

187.5m

187.5m

-1000.0 -1000.0

1000.0

0 0

-500.0 -500.0

500.0 500.0


transformer when the drive on the delta-deltatransformer is at 100% load, the drive on the delta-






La1.I = f...

150.0m

150.0m

200.0m

200.0m

162.5m

162.5m

175.0m

175.0m

187.5m

187.5m

-1000.0 -1000.0

1000.0

0 0

-500.0 -500.0

500.0 500.0



transformer is at 100% load, the drive on the delta-




wye transformer is at 0% load. This is basically the

same as the current we saw in Fig 26.


You can see how the waveform slowly changes

from a 12-pulse shape to a 6-pulse shape.





11

transformer is at 0% load, the drive on the delta-wye transformer is at 100% load. This indicates

how the harmonic current phase angles are shifted

when going through a delta-wye transformer. Thesecondary current of the delta-wye transformer

would look the same as Fig 26 or Fig 37.

La1.I = f...

150.0m

150.0m

200.0m

200.0m

162.5m

162.5m

175.0m

175.0m

187.5m

187.5m

-1000.0 -1000.0

1000.0

0 0

-500.0 -500.0

500.0 500.0

If you were to add the waveform in Fig 38 to thewaveform in Fig 39, you would get the waveform

seen in Fig 34.


La1.I = f...

150.0m

150.0m

200.0m

200.0m

162.5m

162.5m

175.0m

175.0m

187.5m

187.5m

-1000.0 -1000.0

1000.0

0 0

-500.0 -500.0

500.0 500.0

There will always be some 5th and 7th harmonics in

these waveforms since the loads will not be closely

balanced, and because any distortion in the voltagefeeding the two isolation transformers will cause an

imbalance in the currents.

Fig 40 shows the harmonic spectrum of the

waveform seen in Fig 34, which is the best you

could hope for. The waveforms in Fig 38 and Fig

39 will be like the spectrum we saw in Fig 27.Notice that the 11th and 12th harmonics have about

the same magnitude in Fig 40 as in Fig 27.


La1.I = f...

150.0m

150.0m

200.0m

200.0m

162.5m

162.5m

175.0m

175.0m

187.5m

187.5m

-1000.0 -1000.0

1000.0

0 0

-500.0 -500.0

500.0 500.0

Fig 39. Line current with 0% load on delta-deltaand 100% load on delta-wye

0

10

20

30

40

50

60

70

80

90

100

0 6 12 18 24 30 3 6 42 48 54 60 6 6 72 78

Harmonic Number


Fig 40. Harmonic spectrum of the waveform inFig 34. Notice the reduction of the 5th and 7th

harmonic currents due to phase shifting.



12

6. Multi-Pulse Converter Transformer

xfmr% Z

Multi-Phase

Transformer

Drive

DC

AC

AC

DC

DC LinkChoke

3 9M

There are several levels of multi-pulse converter

systems. The most popular are 12-pulse, 18-pulseand 24 pulse. We will concentrate on a well

designed 18-pulse configuration. The heart of amulti-pulse converter is the multi-phase transformeras seen in Fig 41.

Fig 41. Multi-pulse drive configuration utilizing a

multi-phase transformer

Supply Transformer Line Reactor Auto-Transformer 18 Diode Bridge

NOTE: 5 windings per core leg

Supply Transformer Line Reactor Auto-Transformer 18 Diode Bridge

NOTE: 5 windings per core leg

Fig 42. Nine-phase auto-transformer detailalong with the 18-diode bridge creating an 18-pulse converter

Fig 42 shows the details of a 9-phase auto-

transformer used on an 18-pulse drive configuration.Each of the 9 phases from the transformer feed a

pair of diodes in a bridge that contains 18 diodes.

The windings in the transformer are arranged so thatthe taps for the 9 phases are located 40 degrees

apart from each other (9*40 = 360 degrees) and are

located equally distant from a central point. Thisway, the phases are equally spaced in time and have

identical voltage magnitudes.Lsa.I = f...

150.0m

150.0m

200.0m

200.0m

162.5m

162.5m

175.0m

175.0m

187.5m

187.5m

-150.0 -150.0

150.0

0 0

-100.0 -100.0

-50.0 -50.0

50.0 50.0

100.0 100.0

A normal 3-phase transformer has three phases thatare shifted 120 degrees apart from each other

(3*120=360 degrees). Again, with 2 diodes per

phase you get 2 current pulses per phase, and thattotals 6 pulses per cycle. Hence the name 6-pulse

operation. This is clearly seen in Fig 18. Fig 43. Typical line current for an 18-pulse driveconfiguration

The current waveform as seen in Fig 43 shows very

little harmonics. Typically, a well designed 18-pulse converter will have a % Ithd between 4.5 and

6%, creating very little voltage distortion on a

transformer secondary. The harmonic spectrum isseen in Fig 44. Notice that the characteristic or

largest harmonics are the 17th and 19th. This is the

same as the pulse number plus and minus 1 (18

minus 1 = 17, 18 pulse 1 = 19). Similarly, thecharacteristic harmonics for a 6-pulse converter are

the 5th and 7th; the characteristic harmonics for a 12-

pulse converter are the 11th

and 13th

; thecharacteristic harmonics for a 24-pulse converter

are the 23rd and 25th.

0

10

20

30

40

50

60

70

80

90

100

0 6 12 1 8 24 30 36 42 48 54 60 66 72 78

Harmonic Number


Fig 44. Harmonic spectrum for the 18-pulsedrive configuration

DC

ACM

Another method of creating a 9-phase transformer is

to design one with three isolated sets of secondaries.

Each secondary needs to be phase shifted only 20degrees with respect to each other. See Fig 45.

First, you have a delta primary. Then you have a

wye secondary to give you a +30 degree phase shift.

Fig 45. 18-pulse configuration using an isolationtransformer with diode bridges in series



13

Next you have an extended delta for a +10 degreephase shift. And lastly you have another extended

delta for a -10 degree phase shift. Each has to be

wound to provide 1/3 voltage of the rated voltage(for example, 160Vac line-to-line for each

secondary, when used to feed a 480Vac drive).Connect each set to a 6-diode bridge, and connectthe bridges in series with each other. The current

waveform into the delta primary will again be very

close to what is shown in Fig 43. The disadvantage

of this configuration is that each diode needs to besized for the full current rating of the total converter.

The trade-off is that the voltage rating for each

diode is lower. This works well for medium voltageconverter applications and for low hp, low voltage

applications. It becomes too expensive for higher

hp, low voltage drive applications.

0.0

20.0

40.0

60.0

80.0

100.0

120.0

0 25 50 75 100

6p

18p


0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

1.60

1.80

2.00

0 25 50 75 100

6p

18p

0.00

0.20

0.40

0.60

0.80

L a g

L e a d

0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

1.60

1.80

2.00

0 25 50 75 100

6p

18p

0.00

0.20

0.40

0.60

0.80

L a g

L e a d

What is the overall effect of an 18-pulse converter

on a drive and the power grid? Fig 46 to 52 will

give you a comparison to the 6-pulse buffered drivein section 4. The blue traces are for the 6-pulse

drive, the red traces are for the 18-pulse drive. Fig 47. Total Power Factor vs % Load


varies with load. It is fairly linear. The current at100% load is about 101Arms. This current is

composed of about 101Arms of fundamental current

and 5Arms of harmonic current.

90.0

92.0

94.0

96.0

98.0

100.0

102.0

104.0

106.0

108.0

110.0

0 25 50 75 100

6p

18p

Fig 47 shows the total power factor associated with

the line current vs load. At full load, the power

factor is about 0.99 lagging, and slowly drops downto about 0.92 lagging at 10% load.

Fig 48. % Nominal DC Bus Voltage vs % LoadFig 48 shows the change in the DC bus voltage asthe load is changed. 100% of nominal DC bus

voltage is again taken to be 1.35*Vac line-to-line

voltage. For a 480Vac drive, the DC bus voltagefor this 18-pulse drive is about 667Vdc. The slight

boost is due to the placement of the input

connections to the auto-transformer. This workswell for those applications where the line voltage is

a little soft, or where it is necessary to achieve full

power from the motor at full load since the drivehas full voltage on the bus. 0.0

20.0

40.0

60.0

80.0

100.0

120.0

0 20 40 60 80 100

Ithd 6p

Iharm 6p

Ithd 18p

Iharm 18p

Fig 49. % Ithd and Iharm vs % Load



14

Fig 49 shows how Iharm (the harmonic current)varies with load. It also shows the % Ithd as a

dashed line. This is the ratio of the harmonic

current to the fundamental current that is containedin the AC line current.

0.0

2.0

4.0

6.0

8.0

10.0

12.0

14.0

16.0

18.0

20.0

0 25 50 75 100

6p

18p

Fig 50 is a zoomed in detail of % Ithd vs % Load

from Fig 49. As you see, Ithd increases as the loaddecreases due to the different rates of change of the

harmonic current and the fundamental current. Per

IEEE 519, though, the critical point is at full load,which is the minimum point on the % Ithd curve.

Fig 50. Detail of % Ithd vs % Load

0.00

1.00

2.00

3.00

4.005.00

6.00

7.00

8.00

9.00

10.00

0 25 50 75 100

6p

18p

Fig 51 is a zoomed in detail of Iharm vs % Loadfrom Fig 49. It is obvious that as the load decreases,

Iharm also decreases, meaning that the voltage

distortion will be decreasing, too.

Fig 52 is similar to Fig 32, showing how the

displacement power factor and distortion power

factor vary with load. The displacement powerfactor is again quite high throughout the load range.

But since there is so little harmonic current, the

distortion power factor is also very high throughoutthe load range.

Fig 51. Detail of Iharm vs % Load

30.00

40.00

50.00

60.00

70.00

80.00

90.00

100.00

0 10 20 30 40 50 60 70 80 90 100

PF disp

PF dist

PF total

Overall, the 18-pulse converter has very little

impact on the power grid, and is very friendly to the

drive itself. Low harmonics, high power factor,good DC bus voltage at every load point. This is

almost an ideal type of converter.

A 12-pulse converter, on the other hand, would

cause the harmonics to increase 2 to 3 times. Since

there is very little cost difference between and 12-pulse transformer and diode bridge verses an 18-

pulse transformer and diode bridge, and since an

18-pulse converter is able to achieve 5% Ithd at the

input terminals of the drive system, there has been abig drop in demand for 12-pulse converters.


NOTE: There are dozens of transformer andbridge configurations available to achieve 18-pulse operation. Each of these 18-pulseconverters have different characteristics withrespect to line current harmonics, power factor,efficiency, DC bus voltage, etc. The test resultsin this paper refer only to the 18-pulseconfiguration as defined in this section.

A 24-pulse or higher converter is able to furtherreduce the current harmonics, but only by a small

amount. The increased costs provide little payback.

However, due to the large currents involved, thiscould be beneficial for 1500hp drive systems and

larger.



15

7. Passive Filter Transformer

xfmr% Z

Drive

DC

AC

AC

DC

DC LinkChoke

Passive Filter

M

Since the largest harmonic seen in the line currents

of 6-pulse drives is the 5th

harmonic, a passive 5th

harmonic filter added to the line side of a drive is

often a satisfactory solution. See Fig 53.

Some higher performance passive filters are made

with two filtering sections – a 5th and a 7th. As seen

in Fig 54 and 55, a passive filter will significantly

reduce the 5th

harmonic drawn from the transformer.If we examine a passive filter, we see that it is made

up of line reactors and capacitors. The section with

the reactor in series with the capacitor is the tunedsection of the filter. They are usually tuned to a

frequency just below the 5th harmonic, such as at

the 4.7

th

(282Hz). Please realize that a passivefilter designed for a 60Hz power grid will not work

well on a 50Hz power grid due to the different

harmonic frequencies involved.

Fig 53. Drive configuration with the addition of apassive filter

La.I= f(t...

100.0m

100.0m

150.0m

150.0m

110.0m

110.0m

120.0m

120.0m

130.0m

130.0m

140.0m

140.0m

-200.0 -200.0

200.0

0 0

-100.0 -100.0

100.0 100.0

Fig 54. Typical line current for a drive with apassive filter

The reactor between the power distribution system

and the filter helps reduce the resonance effect the

filter will have on the power grid. This helps toisolate the filter from the rest of the power system

so that the tuned section does not become the lowestimpedance point in the whole plant for 5th

harmonics. There are some problems that are

associated with passive filters, though.

0

10

20

30

40

50

60

70

80

90

100

0 6 12 1 8 24 30 36 42 48 54 60 66 72 78

Harmonic Number


a) One is that the capacitors can interact with other

passive filters on the power system, even thougha line reactor is part of the filter assembly. This

can cause voltage and current instabilities.Fig 55. Harmonic spectrum for the drive with a5th harmonic passive filter

b) Along with this, the capacitors draw leadingcurrent from the source. In a plant with several

motor running across the line, this will help

improve the overall power factor. However,there is no control for this. Also, when

operating on back-up generators, the leading

power factor current drawn by the caps cancause unstable operation of the generator.

c) Another problem is that the line reactor willcause a voltage drop in the DC bus when

running at full speed, full load.



16

d) A fourth problem is that the capacitors willcause the DC bus to rise at no load. The

designer of the passive filter has to make several

compromises in order to balance one problemagainst another in order to obtain a best overall

kind of operation.

0.0

20.0

40.0

60.0

80.0

100.0

120.0

0 25 50 75 100

6p

18p

P1

P2

P3

Some passive filters have the ability to disconnectthe filter capacitors from the rest of the filter. This

would be controlled by a contactor that is turned on

or off depending on the speed or load of the drive,or if the drive is operational or not. This helps with

the high DC bus voltage seen at no load and with

generators.


0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

1.60

1.80

2.00

0 25 50 75 100

6p

18p

P1

P2

P3

0.00

0.20

0.40

0.60

0.80

L a g

L e a d

0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

1.60

1.80

2.00

0 25 50 75 100

6p

18p

P1

P2

P3

0.00

0.20

0.40

0.60

0.80

L a g

L e a d

In order to reduce the effect one filter may have on

another, it may be possible to connect more thanone drive to a passive filter output. This reduces thenumber of individual filters on the common

transformer. Also, do not have PF corrections caps

ahead of the passive filter.

The only way to counteract the drop in DC bus

voltage at full speed, full load is to either boost thetransformer secondary or not operate at full speed.

Boosting the transformer secondary has thedisadvantage of causing the DC bus voltage to be

too high under no load conditions. Not operating at

full speed is a more likely operating condition witha drive. If the drive is going to operate at 95% of

full speed or less, then the drop in DC bus voltage

will not be noticed.

Fig 57. Total Power Factor vs % Load

90.0

92.0

94.0

96.0

98.0

100.0

102.0

104.0

106.0

108.0

110.0

0 25 50 75 100

6p

18p

P1

P2

P3 What is the overall effect of a passive filter on a

drive and the power grid? We tested three different

manufacturers’ passive filters with a drive. Fig 56to 61 give you a comparison of the three filters (P1,

P2, P3) to the 6-pulse buffered drive from Section 4

and the 18-pulse drive from Section 6.


0.0

20.0

40.0

60.0

80.0

100.0

120.0

0 20 40 60 80 100

Ithd 6p

Iharm 6p

Ithd 18pIharm 18p

Ithd P1

Iharm P1

Ithd P2

Iharm P2

Ithd P3

Iharm P3


varies with load. It is not linear. The currents at

100% load are similar to the 18-pulse drive.However, as no load is approached, the currents do

not go to zero. Some level off at almost 50% of full

load current! What is causing this? This effect isdue to the capacitor bank that is part of the passive

filter. The capacitors will draw leading power Fig 59. % Ithd and Iharm vs % Load



17

factor current, and is most noticeable at no load.This is saying that when the drive is turned off, you

could have 30 to 50% of full load current flowing to

the passive filter.

0.0

2.0

4.0

6.0

8.0

10.0

12.0

14.0

16.0

18.0

20.0

0 25 50 75 100

6p

18p

P1

P2

P3

Fig 57 shows the total power factor associated withthe line current vs load. At full load, the powerfactor is very close to unity, but then approaches 0.2

to 0.4 leading power factor as the load decreases.

Again, this is due to the cap bank in the filter.Fig 60. Detail of % Ithd vs % Load

Fig 58 shows the change in the DC bus voltage as

the load is changed. 100% of nominal DC bus

voltage is again taken to be 1.35*Vac line-to-linevoltage. At no load, all of the filters provide some

boost in the DC bus voltage due to the cap bank. At

full load, there is a wide range of performance asthe DC bus voltage can be anywhere between 93%

and 102% of nominal. The design of the reactors in

the filter have the biggest impact on this parameter.This can cause the motor to operate at less than

rated V/Hz, causing overheating, if running above

93% of full speed.

0.00

1.00

2.00

3.00

4.005.00

6.00

7.00

8.00

9.00

10.00

0 25 50 75 100

6p

18p

P1

P2

P3


Fig 59 shows how Iharm and % Ithd vary with load.

They are all quite good. For more detail, we havethe next two figures.

Fig 50 is a zoomed in detail of % Ithd vs % Load

from Fig 59. At full load, they are all able to meet

IEEE 519 (which required 8% Ithd). As the load

decreased, the Ithd of the filters would rise veryslowly due to the capacitor bank current discussed

in Fig 56.

Fig 51 is a zoomed in detail of Iharm vs % Load

from Fig 49. This is the real test since this is what

creates the voltage distortion on the transformer. It

is obvious that as the load decreases, Iharm alsodecreases, meaning that the voltage distortion will

be decreasing, too. In the tests, the passive filters

were just a little worse than the 18-pulse drive.

Overall, the passive filters do a good job reducing

the harmonic currents. The biggest drawbacks arethe leading power factor due to the cap bank in the

tuned section of the filter and the drop in DC bus

voltage at full load.



18

8. Active FilterDrive

Transformer

DC

AC

DC LinkChokexfmr

% Z AC

DC

I fu nd I fu nd + Ih ar m

Iharm

Active Filter

AC

DC

M

An active filter is a device that is quite remarkable

in its operation. See Fig 62 for a drive systemutilizing one. It is made up of an inverter bridge

section, like the output inverter section of a drive asshown in Fig 63. However, it is connected such thata cap bank is tied to the DC bus terminals, and the

3-phase terminals are connected to the AC line

through small line reactors and a notch filter. It

then has current sensors that monitor the AC currentgoing to the non-linear load, such as the AC drive.

Fig 62. Drive configuration with the addition ofan active filter

InverterDC to AC

DCBus

Active Filter

A C

L i n e

I n p u t

NotchFilter

fThe key to the operation of an active filter is that itwill supply the harmonic currents required by the

non-linear load. Since these currents have an

average power of zero (zero average watts areconsumed by the non-linear load due to the

harmonic currents), then the harmonic current

supplied by the active filter is absorbed by the drive

during one part of the cycle, and then it is returnedto the filter during the next part of the cycle. The

current simply flows back and forth between the

filter and the drive. It acts like a reservoir forharmonic currents. Most active filters are also able

to supply fundamental reactive current to help withthe displacement power factor.

Fig 63. Details of an active filter showing IGBTinverter section

Ia= f( S,...

-25.00m

-25.00m

24.90m

24.90m

0

0

-20.00m

-20.00m

-10.00m

-10.00m

10.00m

10.00m

20.00m

20.00m

-150.0 -150.0

150.0

0 0

-100.0 -100.0

-50.0 -50.0

50.0 50.0

100.0 100.0

The addition of an active filter to a drive circuitmeans that the transformer does not have to supply

the harmonic current s to the drive. The result is a

clean line current, with just a little ripple due to thecarrier frequency switching of the active filter

inverter section. The current waveform and

spectrum are seen in Fig 64 and 65. The notch filterhelps a great deal with removing the carrier from

the line current. A typical carrier frequency used

for active filters is between 10 and 20kHz.

Fig 64. Typical line current for driveconfiguration with an active filter

0

10

20

30

40

50

60

70

80

90

100

0 6 12 1 8 24 30 36 42 4 8 54 60 66 72 78

Harmonic Number


The effect on the power distribution system is

minimal, similar to an Active Front-End as will be

discussed in Section 9. There is a slight amount of real power consumed due to losses in the filter, but

this is less than ~ 2% of the filter power. The line

current supplied by the transformer will basically be just the fundamental current. The other big

advantage is that one active filter can be associated

with a transformer secondary, supplying whatever

Fig 65. Harmonic spectrum of the line currentfor the drive configuration with an active filter



19

reactive and harmonic currents are needed for all of the loads connected to the transformer. See Fig 66.

In order to meet larger power ratings, active filters

can be paralleled.

AC Drive

DC

AC

DC LinkChoke

AC

DCM

AC Drive

DC

AC

DC LinkChoke

AC

DCM

Transformer

xfmr% Z I fu nd I fu nd + I ha rm

Iharm

Active Filter

AC

DC

DC Drive

AC

DCM

One more advantage is that these units can be addedto a system that is already installed, simply byconnecting the filter to the power lines. No series

reactors are needed (unless the drive does not have

a DC link choke nor an AC line reactor already).

The filter needs only be sized for the harmoniccurrents, not full load current, of the drive. Fig 66. Configuration with multiple drives, AC

and DC, where an active filter can supply all ofthe harmonics in the system

To further improve the utilization of an active filter,the user can use the pseudo-12-pulse technique to

reduce the 5th and 7th harmonics. This will allow

the filter to handle more drives. See Fig 67. If thetotal load were 400A (about 400hp of drives), and if

the Ithd were 30%, this would mean that there is

about 30% of 400A = 120A of harmonic current

drawn from the transformer. The size of the activefilter this system would need is 120A rating.

However, by placing half of the drive hp load on a

delta-wye transformer, then the Ithd would bereduced to about 15%. This would mean that there

is only about 15% of 400A = 60A of harmoniccurrent. The size of the active filter could be

reduced to a 60A rating.

Drive

DC

AC

DC LinkChoke

AC

DCM

Drive

DC

AC

DC LinkChoke

AC

DCM

Transformerxfmr% Z I fu nd I fu nd + I ha rm

Iharm

Active Filter

AC

DC

Fig 67. Combination of an active filter with apseudo 12-pulse system, allowing for a reductionin the current rating of the active filter

It should be noted that active filters employ a

current limiting function as protection for overload

conditions. This can be useful if a slightly smallerfilter can be selected to handle the harmonic load

that is seen 95% of the time. The user would then

allow it to go into current limit during the remainder

of the time. This would not harm the filter, and theonly effect is that the current distortion would

increase during that short time period. If the

voltage distortion remains within the limits then thiswould be an acceptable, lower cost solution.

The only disadvantage is the cost of the active filter.The power electronics required is similar to a drive,

but at about 1/3 of the drive current rating.



20

9. Active Front-End (AFE) Transformer

xfmr% Z

Drive

DC

AC

AC

DC

Notch Filter

M

An active front-end is very similar in construction

to an active filter. The difference is that IGBTs areadded to the diode bridge in the converter section

for the drive. See Fig 68 and 69. This providesseveral advantages for the drive.

Fig 68. Drive configuration with an active front-end as the converter in the drivea) The line current harmonics are greatly reduced,

as long as a filter tuned to the carrier frequency

of the AFE is installed between the powersystem and the input to the AFE. Fig 70 shows

the typical line current, with the harmonic

spectrum shown in Fig 71. If there were nonotch filter, the harmonic spectrum would look

like that shown in Fig 72, where the carrier

frequency of 4kHz appears as side-bands aroundthe 66th harmonic.

A C

L i n e I n p u t

ConverterAC to DC

InverterDC to AC

DCBusFilter

Regenerative AC Drive

A C


b) The AFE is able to operate at unity

displacement power factor from no load to fullload.

Fig 69. Details of a drive utilizing an activeconverter

Lx1.I = ...

145.0

145.0

195.0

195.0

150.0

150.0

162.5

162.5

175.0

175.0

187.5

187.5

00.0 -200.

00.0

0

100.0 -100.

00.0 100.0

c) The AFE is also able to boost the DC busvoltage, providing power ride-through during a

brown-out condition of as much as 50%.

d) The AFE can control current flow while

motoring or regenerating, allowing brakingpower from the drive to be returned to the

power distribution system even during a brown-

out.

Fig 70. Typical line current for the driveconfiguration with an active front-end

In many ways it is like a fully regenerative DC

drive, but with distinct advantages and no

disadvantages to speak of. The only drawback isthe same as the drawback of an active filter – its

cost. It can add about 60% to 90% to the cost of a

drive. However, as you review what all it can dofor you, payback can easily be less than a year in

many applications.0

10

20

30

40

50

60

70

80

90

100

0 6 12 1 8 24 30 36 42 4 8 54 60 66 72 78

Harmonic Number

%

H a r m o n i c C u r r e n

Active Front-End are also known as active

converters, active rectifiers, synchronous converters,

synchronous rectifiers, regenerative converters,regenerative front-ends, and so on.

Fig 71. Harmonic spectrum of the line currentfor the drive configuration utilizing an activefront-end with a notch filter



21

Conceptually, its operation is very simple. TheAFE is like a voltage source whose voltage

magnitude and phase angle with respect to the

power line can be varied. See Fig 73. By changingthe magnitude and phase angle of the voltage

produced by the PWM waveform at the AFEterminals, the magnitude and power factor of thecurrent through the line reactors of the AFE is

regulated from 100% motoring to 0 and all the way

to 100% regenerating. The result of regulating the

line current, the DC bus voltage is controlled asneeded.

0

10

20

30

40

50

60

70

80

90

100

0 6 12 18 24 30 36 42 4 8 54 60 66 72 78

Harmonic Number

%

H a r m o n i c C u r r e n

Fig 72. Harmonic spectrum of the line currentfor the drive configuration utilizing an activefront-end without a notch filter

Fig 74 shows the line-to-neutral voltage in pink, theAFE line-to-neutral voltage (with its PWM

waveform) in blue, and the resulting line current in

red. Notice how the AFE voltage is slightly laggingthe line voltage. This causes the line current to be

at rated current and in phase with the line voltage.

This is at 100% motoring.

Line Current

AFEVoltageSource

LineVoltageSource

Fig 75 again shows the line-to-neutral voltage in

pink, the AFE line-to-neutral voltage (with its PWM

waveform) in blue, and the resulting line current inred. Notice this time how the AFE voltage is

slightly leading the line voltage. This causes theline current to be at rated current and at 180 degrees

out of phase with the line voltage. This is at 100%

regenerating.

Fig 73. Conceptual operation of an active front-end – a PWM voltage source connected to asinusoidal voltage source through a line reactor

If the AFE voltage magnitude and phase with

respect to the line voltage were identical, then theresulting average line current would be zero.

Fig 74. Voltage and current waveforms duringmotoring operation

What is the overall effect of an AFE on a drive and

the power grid? Fig 76 to81 give you a comparisonof an AFE to the 6-pulse drive from Section 4, the

18-pulse drive from Section 6, and the drive with a

passive filter, P1, from Section 7. Please note thatthe line current and harmonic effects are practically

identical for an active filter as described in Section

8.


varies with load. It is quite linear. The currents at100% load are similar to the 18-pulse drive.

However, as no load is approached, the AFE current

heads toward 10A. This is due to the capacitor

Fig 75. Voltage and current waveforms duringregenerating operation



22

bank that is part of the notch filter. Since the capvalues are so low, there is only a small amount of

current at no load.

0.0

20.0

40.0

60.0

80.0

100.0

120.0

0 25 50 75 100

6p

18pP1

AFE

Fig 77 shows the total power factor associated with

the line current vs load. At full load, the powerfactor is very close to unity, but then approaches 0.8

leading power factor as the load decreases. Again,this is due to the cap bank in the notch filter. As the

carrier frequency is increased, the size of the caps

used in the notch filter will decrease, reducing thiseffect.


0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

1.60

1.80

2.00

0 25 50 75 100

6p

18p

P1

AFE

0.00

0.20

0.40

0.60

0.80

L a g

L e a d

0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

1.60

1.80

2.00

0 25 50 75 100

6p

18p

P1

AFE

0.00

0.20

0.40

0.60

0.80

L a g

L e a d

Fig 78 shows the change in the DC bus voltage asthe load is changed. 100% of nominal DC bus

voltage is again taken to be 1.35*Vac line-to-line

voltage. This is one area where an AFE shines.The DC bus voltage is controlled by the AFE micro,even if the line voltage were to dip to 50% of

nominal. During regen, the AFE controller will

sense that the DC bus is increasing, causing it tostart regenerating current to the power lines,

keeping the bus voltage under control at all time. Fig 77. Total Power Factor vs % Load

Fig 79 shows how Iharm and % Ithd vary with load.

The results are very good. For more detail, we havethe next two figures.

90.0

92.0

94.0

96.0

98.0

100.0

102.0

104.0

106.0

108.0

110.0

0 25 50 75 100

6p

18p

P1

AFE

Fig 80 is a zoomed in detail of % Ithd vs % Loadfrom Fig 79. At full load, the AFE is able to easily

meet IEEE 519. As the load decreased, the Ithd of

the AFE rises since the harmonic current does not

change as the fundamental current decreases.

Fig 81 is a zoomed in detail of Iharm vs % Load

from Fig 79. This shows something that is uniqueto active front-ends. The harmonic current is

practically constant at all load levels, and whether it

is motoring or regenerating. All other harmonicmitigation methods have a characteristic where the

harmonic current decreases with load. This is due

to the modulating IGBT bridge.


0.0

20.0

40.0

60.0

80.0

100.0

120.0

0 20 40 60 80 100

Ithd 6p

Iharm 6p

Ithd 18pIharm 18p

Ithd P1

Iharm P1

Ithd AFE

Iharm AFE

Overall, active front-ends do a great job of keeping

the harmonic currents very low with the use of high

carrier frequencies and a good notch filter. Inaddition, it can handle regenerative load currents

and provide very little impact to the power grid Fig 79. % Ithd and Iharm vs % Load



23

system. Its ability to maintain rated DC bus voltageunder brown-out conditions is another plus since it

can keep the process operating during this

disturbance. A drawback is its cost. Compared toan active filter, it is a little less expensive if

integrated with the drive inverter.

0.0

2.0

4.0

6.0

8.0

10.0

12.0

14.0

16.0

18.0

20.0

0 25 50 75 100

6p

18p

P1

AFE

Fig 80. Detail of % Ithd vs % Load

0.00

1.00

2.00

3.00

4.005.00

6.00

7.00

8.00

9.00

10.00

0 25 50 75 100

6p

18p

P1

AFE




24

10. Other Comparisons

0.00

10.00

20.00

30.00

40.00

50.00

60.00

70.00

80.00

90.00

0 25 50 75 100

6p

18p

P1

P2

P3

AFE

How do the efficiencies compare?

Is there a big difference in overall drive systemefficiency when comparing the above methods of

harmonic mitigation? Surprisingly, no! Thedifferences are at the most about 2% at full speed,full load. Not only are they close to one another,

but the one that is a little better at one load level

may become a little worse at another load level.

There is no clear cut advantage that one mitigationmethod has over another when comparing system

efficiency. The following diagram shows how they

compare.

Fig 82. Input kW vs % Load for variousharmonic mitigation solutions

Fig 82 shows the kW into the drive system vs %

Load (measured as torque). Shown are the 6-pulsedrive from Section 4, the 18-pulse drive from

Section 6, three passive filters from Section 7, and

the AFE from Section 9. The drive system is ratedat 100hp. Since the curves are so close to one

another, Fig 83 shows the portion of the chart from

75% load to 100% load only. As you can see, the

AFE takes slightly more watts due to the additionallosses in the converter section and in the notch filter.

The 18-pulse has only about 1.2% more losses dueto the magnetizing current in the multi-phase

transformer. One of the passive filters turned out to

be the most efficient, but not by much.

Fig 84 shows how the system efficiency, including

the motor connected to the drive, varies with % load.For the drive system that is truly the most efficient,

if that is what is absolutely required, the winner is

the basic drive without any filter, without an ACline reactor, without a DC link choke (this is not

shown). However, this gives you the worst

harmonic distortion possible!

50.0

55.0

60.0

65.0

70.0

75.0

80.0

85.0

90.0

95.0

0 25 50 75 100

6p

18p

P1

P2

P3

60.00

65.00

70.00

75.00

80.00

85.00

90.00

75 80 85 90 95 100

6p

18p

P1

P2

P3

AFE

Fig 83. Detail of Input kW vs % Load for variousharmonic mitigation solutions, from Fig 82

Fig 84. % System Efficiency vs % Load forvarious harmonic mitigation solutions

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

10.0

0 25 50 75 100

6p

18pP1

P2

P3

This is why the writer feels that it is more important

to select the harmonic mitigation method that

causes the least amount of problems with the powerdistribution system, with the least detrimental

effects on the drive itself, while still achieving the

IEEE 519 recommended limits.

Remember that the harmonic current limits given in

Table 10.3 refer to the Point of Common Coupling Fig 85. Losses, kW vs % Load for variousharmonic mitigation solutions



25

(at the power meter to the plant), not at the inputterminals of the AC drive. The IEEE 519 standard

never was intended to be an equipment standard.

The most cost effective solution you will want toachieve is to meet the voltage distortion limits given

in Table 10.2 at every point within your plant.

$0

$10,000

$20,000

$30,000

$40,000

$50,000

$60,000

$70,000

$80,000

$90,000

1 0 H

P

2 5 H

P

5 0 H

P

7 5 H

P

1 0 0 H

P

1 5 0 H

P

2 0 0 H

P

2 5 0 H

P

3 0 0 H

P

4 0 0 H

P

5 0 0 H

P

8 0 0 H

P

Active Filter

Passive Filter

18-Pulse

Active Front-End

Fig 85 shows how the system losses, including the

filter or transformer, the drive, and the motor, vary

with load. Notice that the losses are all within 1 to1.5kW. The P3 curve is the passive filter that

would draw about 50% of rated current at no load.

Fig 86. Cost Comparison of Harmonic MitigationSolutions vs HP

How do the costs of the various methods

compare?

This depends on the hp rating you are looking at.

The chart in Fig 86 gives a rough comparison forvarious hp ratings. .

0

5000

10000

15000

20000

25000

30000

35000

40000

6p Drive Passive

Filter 1

Passive

Filter 2

Passiver

Filter 3

Passive

Filter 4

Passive

Filter 5

Active

Filter 1

Active

Filter 2

18p

Drive

As you can see, the 18-pulse solution is more costeffective today than passive filters, active filters, or

active front-ends for drives larger than about 150hp.

Below that hp level, passive filters are more cost

effective. You can also see that active filters aremade in specific, quantum steps. They can be

paralleled in order to achieve larger units. So theyare not optimized for any particular hp rating. The

same is true today for AFEs. The future will bring

the cost of AFEs down where they will be muchcloser to the 18-pulse solution.

Fig 87. Unit Volume (cu in) Comparison ofHarmonic Mitigation Solutions

Fig 87 shows how the volume of each harmonicmitigation solution compares. The 100hp 6-pulse

drive can be used as a basis and is needed for each

of the solutions. There is a large variability inpassive filter sizes.

What are the trends today?

What we are seeing as the trends for today are:

a) 18-pulse drives are firmly established, and are

much more popular than 12-pulse drives since a5% harmonic current distortion can be easily

achieved. This is not possible with 12-pulse

drives. Also, there is no cost advantage of 12-pulse over 18-pulse.



26

b) Passive filters have been improving, with dualstage filters becoming more prevalent.

However, there is still the possibility of

resonance issues among several filters on asingle transformer or on a generator.

c) Active filters are gaining acceptance due to theease of integrating them into an existing system.

Since the cost of iron and copper continue to

increase, and the costs of semiconductors

continue to decrease, this may become morecost effective in the near future.

d) Active front-ends for AC drives is the next stepin the evolution of AC drives. The market will

then have regenerative AC drives, like the

regenerative DC drives that have been sopopular in the workplace. Again, the cost for

the power components will be decreasing.

Integrating the IGBTs into the drive, as opposed

to adding them as an active filter with a drive,will be more cost effective. In addition, as the

switching frequency increases, the notch filter

for the AFE will become smaller and less costly.



27

11. Comparison Chart

Harmonic Mitigation

Solutions Check-List

6 - P u l s

e D r i v

e

1 8

- P u l s

e D r i v

e

P a s s i v

e F i l t e

r

A

c t i v

e F i l t e

r

A c t i v

e F r o n t - E n d

Typical Ithd 20 - 45% 4.5 - 6% 5 - 8% 3 - 5% 3 - 5%

Meet IEEE Special Applications No Yes Marginal Yes Yes

Meet IEEE General Applications No Yes Yes Yes Yes

Meet IEEE Dedicated Applications Yes Yes Yes Yes Yes

Effect of 1% Voltage Unbalance Large Moderate Minimal Minimal Minimal

Potential Low DC Bus No No Yes No No

Potential System Resonance No No Yes No No

Typical Total Power Factor, no / full load 0.75 - 0.95 0.90 - 0.99 0.3 - 1 lead 0.90 - 0.99 0.8 - 1 lead

Efficiency 97% 96.5% 96.5% 96% 96 - 97.5%

Cost Effective Good >150hp <150hp Large Sys Regen, MV

Overall Size (relative to 6-P Drive) 1.0 3.3 2 - 6 3.5 - 5 1.5 - 2.5

Reliability High High Medium Medium Medium

Good

Need to confirm application

May not meet IEEE 519

Notes:

1. The 6-Pulse Drive is one with either a DC Link Choke or with AC Line Reactors.

2. For meeting the various IEEE applications, the drive and mitigation assembly would be the only

item at a Point of Common Coupling. It is possible to have a 6-pulse drive meet IEEE general

applications if there are sufficient linear loads in addition to the converter loads on the transformer.

3. The typical total power factor for a drive with an active front-end, at no load, has much less

leading power factor than that seen with a passive filter. This current is due to the notch filter on the

system.

4. Passive filters do have some drawbacks, but if the system engineer takes these issues into account

in his system design, the likelihood of any problem is minimized.



28

12. Summary Rules-of-Thumb

The following are some rules-of-thumb to use when mitigating line current harmonics with AC Drives:

1. Identify the required PCC and apply techniques most cost effective for that

location.

2. Add a line reactor (or DC link choke if possible) to all unbuffered 6-pulse

drives.

3. Consider use of an active filter on a multiple drive system or MCC lineup tocorrect for harmonic distortion.

4. For an even number of equally sized drives, consider a pseudo 12-pulsesolution by placing half of the load on a phase shifting delta-wye (delta-star)

transformer.

5. Design the system to isolate linear and non-linear loads and create two systems

with 5% and 10% voltage distortion limits.

6. For passive filters on generator power, select a filter with a dropout contactorterminal block for the filter capacitors. This will limit the leading power factor at

no-load and stand-by operation.

7. Take time to understand the benefits and drawbacks of each type of mitigation

solution to assure you meet the requirements of the application and that you canlive with any negative effects created by the chosen harmonic solution.

8. Consider an active front-end if the application requires regenerative operationand harmonic compliance.

9. Perform a preliminary harmonic analysis on your system and explore theeffects of using various harmonic mitigation methods.

10. Never use power factor correction capacitors at the input (or output) of a

drive, or in parallel with passive filters.

These rules should lead you to a successful AC drive application.

Harmonic Mitigation Selection

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